LSZ reduction formula relates the S-matrix element and the time-ordered correlation function, in a complicated equation. However, since $$S=T e^{-i\int d^4x H_I}$$ where $H_I$ is the interaction Hamiltonian, we may compute $\langle f|S|i\rangle$ directly in a perturbative way by expending the exponential in $S$. I think this method is simpler than using the LSZ reduction formula itself, then what is the advantange of the LSZ reduction formula?
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2 Answers
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The premise of the question is that you either use LSZ formula or "directly" compute $\langle f | Te^{-i\int d^4xH_I} |i\rangle$. Implying that these are two unrelated approaches.
While the LSZ formula is derived from $\langle f | S |i\rangle$. You'll just go through the whole derivation of LSZ again if you start from $\langle f | Te^{-i\int d^4xH_I} |i\rangle$. You'll again have to deal with the transition from free states to interacting ones. That'll lead you to the resummation of the terms in your diagrammatic expansion with "dressed" external branches. This will lead to the same external poles structure + renormalization coefficients that the LSZ states.
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2$\begingroup$ Thanks for the answer. It seems that the answer contains some insight related to my question, but I haven't studied neither "dressed external branch" nor "renormalization" yet. Could you explain those terms further or explain to my question in an easier way? $\endgroup$ Commented Sep 25, 2019 at 12:49
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you cannot do some field theory without LSZ reduction formula. Anyway , you just use the result of it by expending that exponantial around the S.
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$\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$ Commented Aug 31 at 13:15